# Math Problem of the Day

$\displaystyle f(x) = \frac{\sin x+x}{x - \cos x}$ $\displaystyle \frac{d}{dx} f(x) = \frac{d}{dx} \left( \frac{\sin x+x}{x - \cos x}\right) = \frac{(\sin x +x )' (x-\cos x) - (\sin x+x)(x-\cos x)'}{(x-\cos x)^2}$ $\displaystyle = \frac{(\cos x+1) (x-\cos x) - (\sin x+x)(1+\sin x)}{(x-\cos x)^2}$ $\displaystyle = \frac{ x\cos x+x - \cos^2 x -\cos x - (\sin^2 x+x\sin x+\sin x+x) }{ (x-\cos x)^2 }$ $\displaystyle = \frac{ x\cos x+x - \cos^2 x -\cos x - \sin^2 x-x\sin x-\sin x-x }{ (x-\cos x)^2 }$ $\displaystyle = \frac{ -\cos^2 x-\sin^2 x+ x\cos x -x\sin x-\sin x-\cos x }{ (x-\cos x)^2 }$ $\displaystyle = \frac{ -1+ x(\cos x -\sin x)-(\sin x+\cos x) }{ (x-\cos x)^2 }$ $\displaystyle f(x) = x^4 \sec x \tan x$ $\displaystyle \frac{d}{dx} \left( x^4 \sec x \tan x \right) = \frac{d}{dx} \left(x^4\right) \sec x \tan x + x^4 \frac{d}{dx} \left(\sec x \right) \tan x + x^4 \sec x \frac{d}{dx} \left(\tan x \right)$ $\displaystyle (fgh)'=f'gh+fg'h+fgh'$ $\displaystyle \frac{d}{dx} \left( x^4 \sec x \tan x \right) = 4x^3 \sec x \tan x + x^4 \frac{d}{dx} \left( \frac{1}{\cos x} \right) \tan x + x^4 \sec x \frac{d}{dx} \left(\frac{\sin x}{\cos x} \right)$ $\displaystyle = 4x^3 \sec x \tan x + x^4 \frac{\sin x}{\cos^2 x} \tan x + x^4 \sec x \frac{1}{\cos^ 2x}$ $\displaystyle = 4x^3 \sec x \tan x + x^4 \sec x \tan^2 x + x^4 \sec^3 x$ $\displaystyle = x^3\sec x \left( 4\tan x + x\tan^2 x+ x\sec^2 x \right)$